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Formatted question description: https://leetcode.ca/all/2243.html

# 2243. Calculate Digit Sum of a String (Easy)

You are given a string `s`

consisting of digits and an integer `k`

.

A **round** can be completed if the length of `s`

is greater than `k`

. In one round, do the following:

**Divide**`s`

into**consecutive groups**of size`k`

such that the first`k`

characters are in the first group, the next`k`

characters are in the second group, and so on.**Note**that the size of the last group can be smaller than`k`

.**Replace**each group of`s`

with a string representing the sum of all its digits. For example,`"346"`

is replaced with`"13"`

because`3 + 4 + 6 = 13`

.**Merge**consecutive groups together to form a new string. If the length of the string is greater than`k`

, repeat from step`1`

.

Return `s`

*after all rounds have been completed*.

**Example 1:**

Input:s = "11111222223", k = 3Output:"135"Explanation:- For the first round, we divide s into groups of size 3: "111", "112", "222", and "23". Then we calculate the digit sum of each group: 1 + 1 + 1 = 3, 1 + 1 + 2 = 4, 2 + 2 + 2 = 6, and 2 + 3 = 5. So, s becomes "3" + "4" + "6" + "5" = "3465" after the first round. - For the second round, we divide s into "346" and "5". Then we calculate the digit sum of each group: 3 + 4 + 6 = 13, 5 = 5. So, s becomes "13" + "5" = "135" after second round. Now, s.length <= k, so we return "135" as the answer.

**Example 2:**

Input:s = "00000000", k = 3Output:"000"Explanation:We divide s into "000", "000", and "00". Then we calculate the digit sum of each group: 0 + 0 + 0 = 0, 0 + 0 + 0 = 0, and 0 + 0 = 0. s becomes "0" + "0" + "0" = "000", whose length is equal to k, so we return "000".

**Constraints:**

`1 <= s.length <= 100`

`2 <= k <= 100`

`s`

consists of digits only.

**Similar Questions**:

## Solution 1.

```
// OJ: https://leetcode.com/problems/calculate-digit-sum-of-a-string/
// Time: O(N)
// Space: O(N)
class Solution {
public:
string digitSum(string s, int k) {
while (s.size() > k) {
string tmp;
int sum = 0;
for (int i = 0; i < s.size(); ++i) {
sum += s[i] - '0';
if ((i + 1) % k == 0 || i == s.size() - 1) {
tmp += to_string(sum);
sum = 0;
}
}
swap(s, tmp);
}
return s;
}
};
```